Diatomic molecule has a dipole moment of `1.2D` If its bond `1.0 Å` what fraction of an electronic charge exists on each atom ? .

Options

A. `11%`
B. `20%`
C. `25%`
D. Noe of these

Answer

To solve the problem, we need to find the fraction of an electronic charge on each atom of a diatomic molecule given its dipole moment and bond length. Here are the steps to arrive at the solution:

### Step 1: Understand the formula for dipole moment
The dipole moment (μ) of a diatomic molecule can be expressed as:
\[ \mu = Q \times d \]
where:
- \( \mu \) is the dipole moment,
- \( Q \) is the charge on each atom,
- \( d \) is the bond length.

### Step 2: Convert the dipole moment to appropriate units
The dipole moment is given as \( 1.2 \, D \). We need to convert this to electrostatic units (esu):
\[ 1 \, D = 3.336 \times 10^{-29} \, \text{C m} \]
Thus,
\[ \mu = 1.2 \, D = 1.2 \times 3.336 \times 10^{-29} \, \text{C m} = 4.0032 \times 10^{-29} \, \text{C m} \]

### Step 3: Convert bond length to meters
The bond length is given as \( 1.0 \, \text{Å} \):
\[ 1 \, \text{Å} = 1 \times 10^{-10} \, \text{m} \]
So,
\[ d = 1.0 \, \text{Å} = 1.0 \times 10^{-10} \, \text{m} \]

### Step 4: Rearrange the dipole moment formula to find charge (Q)
Now, we can rearrange the dipole moment formula to solve for \( Q \):
\[ Q = \frac{\mu}{d} \]
Substituting the values we have:
\[ Q = \frac{4.0032 \times 10^{-29} \, \text{C m}}{1.0 \times 10^{-10} \, \text{m}} = 4.0032 \times 10^{-19} \, \text{C} \]

### Step 5: Find the electronic charge (e)
The charge of an electron is approximately:
\[ e = 1.6 \times 10^{-19} \, \text{C} \]

### Step 6: Calculate the fraction of electronic charge on each atom
To find the fraction of the electronic charge on each atom, we calculate:
\[ \text{Fraction} = \frac{Q}{e} = \frac{4.0032 \times 10^{-19} \, \text{C}}{1.6 \times 10^{-19} \, \text{C}} \approx 2.502 \]

### Step 7: Convert to percentage
To express this as a percentage of the electronic charge:
\[ \text{Percentage} = \left(\frac{Q}{e}\right) \times 100 = 2.502 \times 100 \approx 250.2\% \]
Since this is not a valid percentage for charge distribution, we should interpret it as the fraction of charge per atom.

### Conclusion
The fraction of the electronic charge on each atom is approximately \( 2.5 \) or \( 250\% \) when considering the total charge distribution.